1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Trigo Prob to prove

  1. Nov 19, 2007 #1
    1. The problem statement, all variables and given/known data

    if: [tex] (1+\sqrt{1+a})\tan{\alpha} = (1+\sqrt{1-a}) [/tex]

    Prove that: [tex] \sin{4 \alpha} = a [/tex]

    2. Relevant equations

    [tex] \cos{2\alpha} = 1-2\sin^2{\alpha} [/tex]

    [tex] \tan{\alpha} = \sqrt{\frac{1-\cos{2\alpha}}{1+\cos{2\alpha}}}[/tex]

    3. The attempt at a solution

    We probably have to simplify to express [tex] \tan{\alpha} [/tex] in terms of [tex]\sin{4\alpha}[/tex] and hence need to use the equations given above. Trying componendo and dividendo did no good as it got too complicated.

    [tex]\sin{\alpha}+\sin{\alpha}\sqrt{1+a} = \cos{\alpha}+\cos{\alpha}\sqrt{1-a}[/tex]

    [tex](\sin{\alpha}-\cos{\alpha})^2 = (\cos{\alpha}\sqrt{1-a}-\sin{\alpha}\sqrt{1+a})^2 [/tex]

    [tex]1-\sin{2\alpha}=\cos^2{\alpha}(1-a)+\sin^2{\alpha}(1+a)-\sin{2\alpha}\sqrt{1-a^2}[/tex]

    Though I could simplify after this, I cannot understand how to get [tex]\sin{4\alpha} [/tex] out of it. What is looking even more difficult is how to remove the [tex] \sqrt{1-a^2} [/tex] term.

    I would be grateful if somebody could guide me in solving this.
     
  2. jcsd
  3. Nov 19, 2007 #2
    Wait, is the [tex] \sqrt{1-a^2} [/tex] inside of your last sine term or is it a scalar multiplying your last sine term?
     
  4. Nov 20, 2007 #3
    No, it is a scalar multiplying the last sine term.
     
    Last edited: Nov 20, 2007
  5. Nov 20, 2007 #4
    Looks like I got it; beginning from where I left,

    [tex]1-\sin{2\alpha} = \cos^2{\alpha-a\cos^2{\alpha}+\sin^2{\alpha}+a\sin^2{\alpha}-(\sin{2\alpha})(\sqrt{1-a^2})[/tex]

    -After taking 'a' common...

    [tex]Since, \cos^2{\alpha}+\sin^2{\alpha}=1[/tex]

    [tex]\Rightarrow\sin{2\alpha}=a\cos{2\alpha}+(\sin{2\alpha})(\sqrt{1-a^2})[/tex]

    [tex]\Rightarrow(\sin{2\alpha}-a\cos{2\alpha})^2=(\sqrt{1-a^2}.\sin{2\alpha})^2[/tex]

    [tex]\Rightarrow\sin^2{2\alpha}+a^2\cos^2{2\alpha}-a\sin{4\alpha}=\sin^2{2\alpha}-a^2\sin^2{2\alpha}[/tex]

    the term [tex]sin^2{2\alpha}[/tex] gets cancelled on both sides and taking 'a' common we get>>

    [tex]\Rightarrow \sin{4\alpha}=a[/tex]

    VOILA!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Trigo Prob to prove
  1. Trigo doubt (Replies: 4)

  2. A bit of Trigo (Replies: 1)

  3. Trigo/Geom. problem (Replies: 7)

  4. Limit prob (Replies: 4)

Loading...